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प्रश्न
Solve \[\left| x + 1 \right| + \left| x \right| > 3\]
उत्तर
\[\text{ We have }, \left| x + 1 \right| + \left| x \right| > 3\]
\[\text{ As }, \left| x + 1 \right| = \binom{\left( x + 1 \right), x \geq - 1}{ - \left( x + 1 \right), x < - 1}\]
\[\text{ and } \left| x \right| = \binom{x, x \geq 0}{ - x, x < 0}\]
\[\text{ Case I: When } x < - 1, \]
\[\left| x + 1 \right| + \left| x \right| > 3\]
\[ \Rightarrow - \left( x + 1 \right) - x > 3\]
\[ \Rightarrow - 2x - 1 > 3\]
\[ \Rightarrow - 2x > 4\]
\[ \Rightarrow x < \frac{4}{- 2}\]
\[ \Rightarrow x < - 2\]
\[\text{ So }, x \in \left( - \infty , - 2 \right)\]
\[\text{ Case II: When } - 1 \leq x < 0, \]
\[\left| x + 1 \right| + \left| x \right| > 3\]
\[ \Rightarrow \left( x + 1 \right) - x > 3\]
\[ \Rightarrow 1 > 3, \text{ which is not possible }\]
\[\text{ So }, x \in \phi\]
\[\text{ Case III: When x } \geq 0, \]
\[\left| x + 1 \right| + \left| x \right| > 3\]
\[ \Rightarrow \left( x + 1 \right) + x > 3\]
\[ \Rightarrow 2x + 1 > 3\]
\[ \Rightarrow 2x > 2\]
\[ \Rightarrow x > \frac{2}{2}\]
\[ \Rightarrow x > 1\]
\[\text{ So }, x \in \left( 1, \infty \right)\]
\[\text{ From all the cases, we get }\]
\[x \in \left( - \infty , - 2 \right) \cup \left( 1, \infty \right)\]
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